Laws of Motion — Page 3 | JEE Physics

Q21

Two forces P and Q, of magnitude 2F and 3F, respectively, are at an angle

\theta

with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle

\theta

is -

A 90o

B 60o

C 30o

D 120o

Correct Answer

Option D

Solution

4F2 + 9F2 + 12F2 cos $\theta$ = R2 4F2 + 36F2 + 24F2 cos $\theta$ = 4R2 4F2 + 36F2 + 24F2 cos $\theta$ = 4(13F2 + 12F2cos $\theta$ ) = 52F2 + 48F2cos $\theta$ cos $\theta$ = $-$

{{12{F^2}} \over {24{F^2}}}

= $-$

{1 \over 2}

Q22

A block of mass

5 \mathrm{~kg}

is placed at rest on a table of rough surface. Now, if a force of

30 \mathrm{~N}

is applied in the direction parallel to surface of the table, the block slides through a distance of

50 \mathrm{~m}

in an interval of time

10 \mathrm{~s}

. Coefficient of kinetic friction is (given,

g=10 \mathrm{~ms}^{-2}

):

A 0.25

B 0.75

C 0.60

D 0.50

Correct Answer

Option D

Solution

\begin{aligned} & S=u t+\frac{1}{2} a t^2 \\\\ & 50=0+\frac{1}{2} \times a \times 100 \\\\ & a=1 \mathrm{~m} / \mathrm{s}^2 \\\\ & F-\mu m g=m a \\\\ & 30-\mu \times 50=5 \times 1 \\\\ & 50 \mu=25 \\\\ & \mu=\frac{1}{2} \end{aligned}

Q23

A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate

{{dM\left( t \right)} \over {dt}}

= bv2(t), where v(t) is its instantaneous velocity. The instantaneous acceleration of the satellite is :

A -bv3(t)

B

- {{2b{v^3}} \over {M\left( t \right)}}

C

- {{b{v^3}} \over {M\left( t \right)}}

D

- {{b{v^3}} \over {2M\left( t \right)}}

Correct Answer

Option C

Solution

Given

{{dM\left( t \right)} \over {dt}}

= bv2(t) In free space no external force so there in only thrust force on rocket. Fthrust = v

{{dm} \over {dt}}

Force on satellite =

- \overrightarrow v {{dm\left( t \right)} \over {dt}}

M(t)a = – v (bv2) $\Rightarrow$ a =

- {{b{v^3}} \over {M\left( t \right)}}

Q24

An insect is at the bottom of a hemispherical ditch of radius 1 m. It crawls up the ditch but starts slipping after it is at height h from the bottom. If the coefficient of friction between the ground and the insect is 0.75, then h is : (g = 10 ms–2)

A 0.45 m

B 0.60 m

C 0.20 m

D 0.80 m

Correct Answer

Option C

Solution

For balancing mgsin $\theta$ = f mgsin $\theta$ = $\mu$ mgcos $\theta$ $\Rightarrow$ tan $\theta$ = $\mu$ =

{3 \over 4}

h = R – R cos $\theta$ = R - R

\left( {{4 \over 5}} \right)

=

{R \over 5}

$\therefore$ h =

{R \over 5}

=

{1 \over 5}

= 0.2 m

Q25

A particle moving in the xy plane experiences a velocity dependent force

\overrightarrow F = k\left( {{v_y}\widehat i + {v_x}\widehat j} \right)

, where vx and vy are the x and y components of its velocity

\overrightarrow v

. If

\overrightarrow a

is the acceleration of the particle, then which of the following statements is true for the particle?

A kinetic energy of particle is constant in time

B quantity

\overrightarrow v \times \overrightarrow a

is constant in time

C quantity

\overrightarrow v .\overrightarrow a

is constant in time

D

\overrightarrow F

arises due to a magnetic field

Correct Answer

Option B

Solution

Given

\overrightarrow F = k\left( {{v_y}\widehat i + {v_x}\widehat j} \right)

$\Rightarrow$ m

\overrightarrow a

=

k\left( {{v_y}\widehat i + {v_x}\widehat j} \right)

$\Rightarrow$

\overrightarrow a = {k \over m}\left( {{v_y}\widehat i + {v_x}\widehat j} \right)

Also

{{d{v_x}} \over {dt}} = {k \over m}{v_x}

and

{{d{v_y}} \over {dt}} = {k \over m}{v_y}

{{d{v_x}} \over {d{v_y}}} = {{{v_y}} \over {{v_x}}}

$\Rightarrow$

\int {{v_x}d{v_x}} = \int {{v_y}} d{v_y}

$\Rightarrow$

v_y^2 = v_x^2 + C

$\Rightarrow$

v_y^2 - v_x^2 = C

= Constant From Option (B),

\overrightarrow v \times \overrightarrow a

=

\left( {{v_x}\widehat i + {v_y}\widehat j} \right) \times {k \over m}\left( {{v_y}\widehat i + {v_x}\widehat j} \right)

=

\left( {v_x^2\widehat k - v_y^2\widehat k} \right) \times {k \over m}

=

\left( {v_x^2 - v_y^2} \right) \times {k \over m}\widehat k

= Constant

Q26

A particle is projected with velocity v0 along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. ma =

-

\alpha

x2. The distance at which the particle stops :

A

{\left[ {{{3mv_0^2} \over {2\alpha }}} \right]^{{1 \over 3}}}

B

{\left( {{{2{v_0}} \over {3\alpha }}} \right)^{{1 \over 3}}}

C

{\left( {{{3v_0^2} \over {2\alpha }}} \right)^{{1 \over 2}}}

D

{\left( {{{2v_0^2} \over {3\alpha }}} \right)^{{1 \over 2}}}

Correct Answer

Option A

Solution

Given, speed of projection = v0 Damping force, F = ma = $-$ $\alpha$ x2 $\Rightarrow$ a = $-$ $\alpha$ x2 / m Also,

a = v{{dv} \over {dx}}

\Rightarrow vdv = a\,dx = - {\alpha \over m}{x^2}dx

Integrating both sides, we get

\int_{{v_0}}^v {vdv = \int_0^x { - {\alpha \over m}{x^2}dx} }

\Rightarrow \left( {{{{v^2}} \over 2}} \right)_{{v_0}}^0 = - {\alpha \over m}\left( {{{{x^3}} \over 3}} \right)_0^x

\Rightarrow 0 - v_0^2/2 = - {\alpha \over m}{{{x^3}} \over 3} \Rightarrow x = {\left( {{{3m} \over 2}{{v_0^2} \over \alpha }} \right)^{1/3}}

Q27

A particle of mass M originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation

F = {F_0}\left[ {1 - {{\left( {{{t - T} \over T}} \right)}^2}} \right]

Where F0 and T are constants. The force acts only for the time interval 2T. The velocity v of the particle after time 2T is :

A 2F0T/M

B F0T/2M

C 4F0T/3M

D F0T/3M

Correct Answer

Option C

Solution

At t = 0, u = 0

a = {{{F_0}} \over M} - {{{F_0}} \over {M{T^2}}}{(t - T)^2} = {{dv} \over {dt}}

\int\limits_0^v {dv = \int\limits_{t = 0}^{2T} {\left( {{{{F_0}} \over M} - {{{F_0}} \over {M{T^2}}}{{(t - T)}^2}} \right)dt} }

V = \left[ {{{{F_0}} \over M}t} \right]_0^{2T} - {{{F_0}} \over {M{T^2}}}\left[ {{{{t^3}} \over 3} - {t^2}T + {T^2}t} \right]_0^{2T}

$\Rightarrow$

V = {{4{F_0}T} \over {3M}}

Q28

A block of mass 2 kg moving on a horizontal surface with speed of 4 ms

-

1 enters a rough surface ranging from x = 0.5 m to x = 1.5 m. The retarding force in this range of rough surface is related to distance by F =

-

kx where k = 12 Nm

-

1. The speed of the block as it just crosses the rough surface will be :

A zero

B 1.5 ms

-

1

C 2.0 ms

-

1

D 2.5 ms

-

1

Correct Answer

Option C

Solution

F = - 12x

mv{{dv} \over {dx}} = - 12x

\int_4^v {vdv = - 6\int_{0.5}^{1.5} {xdx} }

(

m = 2

kg)

{{{v^2} - 16} \over 2} = - 6\left[ {{{{{1.5}^2} - {{0.5}^2}} \over 2}} \right]

{{{v^2} - 16} \over 2} = - 6

v = 2

m/sec

Q29

A person is standing in an elevator. In which situation, he experiences weight loss?

A When the elevator moves upward with constant acceleration

B When the elevator moves downward with constant acceleration

C When the elevator moves upward with uniform velocity

D When the elevator moves downward with uniform velocity

Correct Answer

Option B

Solution

N1 = mg N2 = mg + ma N3 = mg $-$ ma N4 = mg N5 = mg

Q30

An object of mass 5 kg is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of 10 N throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use g = 10 ms

-

2].

A 1 : 1

B

\sqrt 2

:

\sqrt 3

C

\sqrt 3

:

\sqrt 2

D 2 : 3

Correct Answer

Option B

Solution

Let time taken to ascent is t1 and that to descent is t2. Height will be same so

H = {1 \over 2} \times 12t_1^2 = {1 \over 2}\times8t_2^2

\Rightarrow {{{t_1}} \over {{t_1}}} = {{\sqrt 2 } \over {\sqrt 3 }}